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A Proton Having a Speed Of 3.0×106 m/s3.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }

question 6

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A proton having a speed of 3.0×106 m/s3.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s } in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m0.20 \mathrm {~m} within the field. What is the magnitude of the magnetic field? (e( e =1.60×1019C,mproton =1.67×1027 kg= 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} )


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